2016 AMC 10B Problems/Problem 2

Problem

If $n\heartsuit m=n^3m^2$ , what is $\frac{2\heartsuit 4}{4\heartsuit 2}$ ?

$\textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 4$

Solution 1

$\frac{2^3(2^2)^2}{(2^2)^32^2}=\frac{2^7}{2^8}=\frac12$ which is $\textbf{(B)}$ .

Solution 2

We can replace $2$ and $4$ with $a$ and $b$ respectively. Then substituting with $n$ and $m$ we can get $\dfrac{a^3b^2}{b^3a^2}=\dfrac{a}{b}$ and substitute to get $\dfrac{2}{4}=\boxed{\dfrac{1}{2}}$ which is $\boxed{\textbf{(B)}}$

Video Solution (CREATIVE THINKING)

https://youtu.be/boCvD0Hb6h0

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Video Solution

https://youtu.be/W7IwD6sZWco

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2016 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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